A Mathematical Model of Granule Cell Generation During Mouse Cerebellum Development

Abstract

Determining the cellular basis of brain growth is an important problem in developmental neurobiology. In the mammalian brain, the cerebellum is particularly amenable to studies of growth because it contains only a few cell types, including the granule cells, which are the most numerous neuronal subtype. Furthermore, in the mouse cerebellum granule cells are generated from granule cell precursors (gcps) in the external granule layer (EGL), from 1 day before birth until about 2 weeks of age. The complexity of the underlying cellular processes (multiple cell behaviors, three spatial dimensions, time-dependent changes) requires a quantitative framework to be fully understood. In this paper, a differential equation-based model is presented, which can be used to estimate temporal changes in granule cell numbers in the EGL. The model includes the proliferation of gcps and their differentiation into granule cells, as well as the process by which granule cells leave the EGL. Parameters describing these biological processes were derived from fitting the model to histological data. This mathematical model should be useful for understanding altered gcp and granule cell behaviors in mouse mutants with abnormal cerebellar development and cerebellar cancers.

Appendix: Derivation of the Formula for Average Clone Size for the Linear \(\delta (t)\) Probability Function

The total number of granule cells generated is given by:

$$\begin{aligned} \hbox {N}_\mathrm{g} (\infty )=\hbox {N}_\mathrm{o} (0)+\alpha _\mathrm{P} \int _0^\infty \hbox {N}_\mathrm{o} (t) \hbox {d}t \end{aligned}$$

(13)

where

$$\begin{aligned} \hbox {N}_\mathrm{o} (t)=\hbox {N}_\mathrm{o} (0) \hbox {e}^{\alpha _\mathrm{P} \int _0^t \left( {1-2\delta \left( \tau \right) } \right) \hbox {d}\tau } \end{aligned}$$

(14)

and

$$\begin{aligned} \delta \left( \tau \right) =\left\{ {{\begin{array}{ll} \tau /T, &{} 0\le \tau <T \\ 1, &{} \tau \ge T \\ \end{array} }} \right. \end{aligned}$$

(15)

Substituting Eqs. (14) and (15) into (13), we have:

$$\begin{aligned} \frac{N_\mathrm{g} (\infty )}{N_\mathrm{o} (0)}=1+\alpha _\mathrm{P} \int _0^\mathrm{T} \hbox {e}^{\alpha _\mathrm{P} \int _0^t \left( {1-\frac{2\tau }{T}} \right) \hbox {d}\tau } \hbox {d}t+\alpha _\mathrm{P} \int _T^\infty \hbox {e}^{-\alpha _\mathrm{P} \left[ {\int _0^\mathrm{T} \left( {1-\frac{2\tau }{T}} \right) \hbox {d}\tau + \int _\mathrm{T}^t \hbox {d}\tau } \right] }\hbox {d}t\quad \end{aligned}$$

(16)

Note that $\int _0^T \left( {1-\frac{2\tau }{T}} \right) \hbox {d}\tau =T-\frac{T^{2}}{T}=0$, so Eq. (16) becomes:

$$\begin{aligned} \frac{N_\mathrm{g} (\infty )}{N_\mathrm{o} (0)}= & {} 1+\alpha _\mathrm{P} \int _0^T \hbox {e}^{\alpha _\mathrm{P} \left( {t-\frac{t^{2}}{T}} \right) } \hbox {d}t+\alpha _\mathrm{P} \int _\mathrm{T}^\infty \hbox {e}^{-\alpha _\mathrm{P} \left( {t-T} \right) } \hbox {d}t\nonumber \\= & {} 2+\alpha _\mathrm{P} \int _0^T \hbox {e}^{\alpha _\mathrm{P} \left( {t-\frac{t^{2}}{T}} \right) } \hbox {d}t \end{aligned}$$

(17)

By expressing

$$\begin{aligned} \left( {t-\frac{t^{2}}{T}} \right) = \frac{T}{4}-\frac{1}{T}\left( {t^{2}-tT+ \frac{T^{2}}{4}} \right) = \frac{T}{4}-\frac{1}{T}\left( {t- \frac{T}{2}} \right) ^{2}, \end{aligned}$$

Equation (17) can be rewritten as:

$$\begin{aligned} \frac{\hbox {N}_\mathrm{g} (\infty )}{\hbox {N}_\mathrm{o} (0)}=2+\alpha _\mathrm{P} \hbox {e}^{\left( {\frac{\alpha _\mathrm{P} T}{4}} \right) }\int _0^T \hbox {e}^{-\frac{\alpha _\mathrm{P}}{T} \left( {t-\frac{T}{2}} \right) ^{2}}\hbox {d}t \end{aligned}$$

(18)

Now, we will define:

$$\begin{aligned} v=\sqrt{\frac{\alpha _\mathrm{P} }{T}}\left( {t-\frac{T}{2}} \right) ,\hbox { so that }\hbox {d}v=\sqrt{\frac{\alpha _\mathrm{P} }{T}}\hbox {d}t \end{aligned}$$

(19)

Then:

$$\begin{aligned} t=0 \Rightarrow v=-\frac{\sqrt{\alpha _\mathrm{P} T}}{2};\hbox { and }t=T \Rightarrow v=\frac{\sqrt{\alpha _\mathrm{P} T}}{2} \end{aligned}$$

(20)

Substituting Eqs. (19) and (20) into (18), we have:

$$\begin{aligned} \frac{N_\mathrm{g} (\infty )}{N_\mathrm{o} (0)}= & {} 2+\sqrt{\alpha _\mathrm{P} T}\hbox {e}^{\left( {\frac{\alpha _\mathrm{P} T}{4}} \right) }\int _{- \frac{\sqrt{\alpha _\mathrm{P} T}}{2}}^{\frac{\sqrt{\alpha _\mathrm{P} T}}{2}} \hbox {e}^{-v^{2}}\hbox {d}v\\= & {} 2+2\sqrt{\alpha _\mathrm{P} T}\hbox {e}^{\left( {\frac{\alpha _\mathrm{P} T}{4}} \right) }\int _0^{\frac{\sqrt{\alpha _\mathrm{P} T}}{2}} \hbox {e}^{-v^{2}}\hbox {d}v \end{aligned}$$

Therefore the average clone size is:

$$\begin{aligned} \frac{N_\mathrm{g} (\infty )}{N_\mathrm{o} (0)}=2+\sqrt{\left( {\alpha _\mathrm{P} T} \right) \pi } \hbox {e}^{\left( {\frac{\alpha _\mathrm{P} T}{4}} \right) } \hbox {erf}\left( {\frac{\sqrt{\alpha _\mathrm{P} T}}{2}} \right) \end{aligned}$$

(21)

where $\hbox {erf}(x)$ is the error function defined as (Abramowitz and Stegun 1964):

$$\begin{aligned} \hbox {erf}(x)= \frac{2}{\sqrt{\pi }} \int _0^x \hbox {e}^{-v^{2}}\hbox {d}v \end{aligned}$$

(22)